23.0 g KCl 0.700 L 58.4 mL KCl 42.0 mL AgNO3

A solution is prepared by placing 23.0 g of KCl in a 0.700 L volumetric flask

and adding water to dissolve the solid, then filling the flask to the mark.

What is the molarity of an AgNO3 solution

if 58.4 mL of the KCl solution react exactly with 42.0 mL of the AgNO3 solution?

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KCl의 몰질량 = 74.55 g/mol 이므로,

23.0 g / (74.55 g/mol) = 0.30852 mol KCl

KCl 용액의 몰농도를 계산하면,

몰농도 = 용질 mol수 / 용액 L수

( 참고 https://ywpop.tistory.com/3222 )

= 0.30852 mol / 0.700 L

= 0.441 M

KCl + AgNO3 → AgCl + KNO3

KCl : AgNO3 = 1 : 1 계수비(= 몰수비) 이므로,

(0.441 M) (58.4 mL) = (? M) (42.0 mL)

? = (0.441) (58.4) / (42.0) = 0.613 M

답: 0.613 M

[키워드] KCl + AgNO3 기준, AgNO3 + KCl 기준, 화학량론